First Order ODE/y' = (x + y)^2

Theorem

The first order ODE:

$\dfrac {\d y} {\d x} = \paren {x + y}^2$

has the general solution:

$x + y = \map \tan {x + C}$

Proof

Make the substitution:

$z = x + y$

Then from First Order ODE in form $y' = f (a x + b y + c)$ with $a = b = 1$:

\(\ds x\)

\(=\)

\(\ds \int \frac {\d z} {z^2 + 1}\)

\(\ds \)

\(=\)

\(\ds \arctan z + C_1\)

Primitive of $\dfrac 1 {x^2 + a^2}$

\(\ds \leadsto \ \ \)

\(\ds y - C_1\)

\(=\)

\(\ds \map \arctan {x + y}\)

\(\ds \leadsto \ \ \)

\(\ds x + y\)

\(=\)

\(\ds \map \tan {x + C}\)

letting $C = - C_1$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.7$: Homogeneous Equations: Problem $3 \ \text{(a)}$